题:https://leetcode.com/problems/two-sum/description/
题目
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
思路
1.扫描一遍nums [i] ,将nusm[i ],i 映射到map,时间复杂度为O(logN),空间复杂度为O(N);
2.在map中,扫描一遍target - nums [i],若存在且返回的下标不为i。返回 i 与 上述下标。
笔记
1.map
声明&定义:
map<int,int> tmap;
查找:
map<int,int>::iterator findres = tmap.find(key);
if(findres != tmap.end()) .....(查找到)
findres->first; //map.key
findres->second; //map.value
Code
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
int ibegin = 0,iend = nums.size()-1;
map<int,int> maprec;
vector<int> res;
for(int i = 0 ;i<nums.size() ; i++) maprec[nums[i]] = i;
for(int i = 0 ;i<nums.size() ; i++){
int complement = target - nums[i];
map<int, int>::iterator findres= maprec.find(complement);
if(findres != maprec.end() && findres->second != i){
res.push_back(i);
res.push_back(findres->second);
return res;
}
}
}
};